# Solve for f(x) the integral equation int_1^xf(t)dt=x(f(x))^2 ?

Nov 19, 2016

$f \left(x\right) = 1 - \frac{1}{\sqrt{x}}$ and $f \left(x\right) = 0$

#### Explanation:

${\int}_{1}^{x} f \left(t\right) \mathrm{dt} = x {\left(f \left(x\right)\right)}^{2}$ deriving both sides

$f \left(x\right) = {\left(f \left(x\right)\right)}^{2} + 2 x f \left(x\right) f ' \left(x\right)$ or

$\left(2 x f ' \left(x\right) + f \left(x\right) - 1\right) f \left(x\right) = 0$ Supposing that $f \left(x\right) \ne 0$ we have

$2 x f ' \left(x\right) + f \left(x\right) - 1 = 0$ Solving this linear differential equation we obtain

$f \left(x\right) = 1 + {C}_{1} / \sqrt{x}$ The constant is determined submiting this solution into the first relationship giving

$\frac{x + 2 \left(\sqrt{x} - 1\right) {C}_{1} - 1}{x} = {\left(1 + {C}_{1} / \sqrt{x}\right)}^{2}$ giving ${C}_{1} = - 1$ so the function is

$f \left(x\right) = 1 - \frac{1}{\sqrt{x}}$ and also $f \left(x\right) = 0$